Question

A plane square sheet of charge of side $0.5\, m$ has uniform surface charge density. An electron at $1\, cm$ from the centre of the sheet experiences a force of $1.6 \times 10^{-12}\, N$ directed away from the sheet. The total charge on the plane square sheet is $\left(\varepsilon_{0}=8.854 \times 10^{-12} C ^{2} m ^{-2}\, N ^{-1}\right)$

• A. $16.25 \,\mu C$
• B. $ - 22.15 \,\mu C$
• C. $-44.27 \,\mu C$
• D. $ 144.27 \,\mu C $
• E. $ 8.854 \,\mu C$

Answer 1

Answer: (C)

Given, $l=0.5 \,m$

$F=1.6 \times 10^{-12}\, N$
$q=-1.6 \times 10^{-19} C \,\,\,$ (for electron)
$\varepsilon_{0}=8.854 \times 10^{-12} \,C ^{2} \,m ^{-2} \,N ^{-1}$
Electric field $E=\frac{F}{q}$
or $F=q E \,\,\,...(i)$
Electric field due to a square plane sheet of charge
$E=\frac{\sigma}{2 \varepsilon_{0}}$
where $\sigma=$ surface charge density
$\varepsilon_{0}=$ permittivity of free space
Putting the value of $E$ in Eq. (i), we get
$ F =q \cdot \frac{\sigma}{2 \varepsilon_{0}} $
$\sigma =2 \varepsilon_{0} \cdot \frac{F}{q}$
$\sigma=\frac{2 \times 8.854 \times 10^{-12} \times 1.6 \times 10^{-12}}{-1.6 \times 10^{-19}}$
$ \sigma =-17.708 \times 10^{-5} $
$ \sigma A =l^{2} \times\left(-17.708 \times 10^{-5}\right) $
$=0.5 \times 0.5 \times\left(-17.708 \times 10^{-5}\right)$
$=-4.427 \times 10^{-5}$
$=-44.27 \times 10^{-6}\, C$
$=-44.27 \,\mu \,C$